Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.awt.Point;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.*;
import java.util.*;

public class Main {
	//////
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	//////
	public static long modularExponentiation(long x, long n, long M) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x) % M;
			x = (x * x) % M;
			n = n / 2;
		}
		return result;
	}

	public static long modularExponentiation1(long x, long n) {
		long result = 1;
		while (n > 0) {
			if (n % 2 == 1)
				result = (result * x);
			x = (x * x);
			n = n / 2;
		}
		return result;
	}

	/////
	static long d(int x, int y) {
		if (Math.abs(x - y) > 1) {
			return y - x;
		} else
			return 0;

	}

	static int closestSquare(int num) {

		if (num == 0) {
			return -1;
		}
		int root = (int) Math.sqrt(num);
		int another = root + 1;
		root = root * root;
		another = another * another;
		if (another - num > num - root) {
			return num - root;
		} else
			return another - num;

	}

	
	public static void main(String[] args) throws IOException {

		FastReader scan = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);

		int N=scan.nextInt();
		int M=scan.nextInt();
		int[][] A=new int[N][M];
		int[][] temp=new int[N][M];
		for(int i=0;i<N;i++){
			for(int j=0;j<M;j++){
				A[i][j]=scan.nextInt();
			}
		}
		int ansx=0,ansy=0;int flagg=0;
		for(int i=0;i<N;i++){
			for(int j=0;j<M;j++){
				if(A[i][j]==0){
					flagg=1;
					ansx=i;ansy=j;
				}
			}
		}
		int count=0;
		ArrayList<String> arr=new ArrayList<String>();
		
		while(flagg==0){
			if(N<M){
			for(int i=0;i<M;i++){
				A[0][i]--;
				
			}
			count++;
			arr.add("row 1");
			for(int i=0;i<M;i++){
				if(A[0][i]==0){
					flagg=1;
					ansx=0;ansy=i;
				}
			}
			}else{
				for(int i=0;i<N;i++){
					A[i][0]--;
					
				}
				count++;
				arr.add("col 1");
				for(int i=0;i<N;i++){
					if(A[i][0]==0){
						flagg=1;
						ansx=i;ansy=0;
					}
				}
			}
		}
		
		for(int i=0;i<N;i++){
			if(ansx!=i){
				for(int k=0;k<A[i][ansy];k++){
				count++;
				for(int l=0;l<M;l++){
					temp[i][l]++;
				}
				arr.add("row "+(i+1));
				}
			}
		}
		for(int i=0;i<M;i++){
			if(ansy!=i){
				for(int k=0;k<A[ansx][i];k++){
				count++;
				for(int l=0;l<N;l++){
					temp[l][i]++;
				}
				arr.add("col "+(i+1));
				}
			}
		}
		int flag=0;
		for(int i=0;i<N;i++){
			for(int j=0;j<M;j++){
				//System.out.print(temp[i][j]);
				if(A[i][j]!=temp[i][j]){
					flag=1;
				}
			}
			//System.out.println();
		}
		
		if(flag==1){
			System.out.println("-1");
			return;
		}else
		{
			System.out.println(count);
			for(int i=0;i<arr.size();i++){
				System.out.println(arr.get(i));
			}
		}
		
		
		
	}
}